599 N skier begins on a hill 32 m above the valley floor. He travels down to the valley floor and up a 18 m hill on the other si
1 answer:
Answer:
v = 129.5 m / s
Explanation:
For this exercise we must use the conservation of mechanical energy.
Starting point. On the side of the first hill
Em₀ = U = m g h₁
Final point. On the other hill
Em_f = K + U = ½ m v² + mgh₂
where h2 is the latura of the ora hill
as they indicate that there is no friction, energy is conserved
Em₀ = Em_f
mgh₁ = ½ mv² + m gh₂
v² = 2 mg (h1-h2)
v =
let's calculate
v =
v = 129.5 m / s
This speed is horizontal at the top of the hill
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Answer:
Ein: 2.75*10^-3 N/C
Explanation:
The induced electric field can be calculated by using the following path integral:
Where:
dl: diferencial of circumference of the ring
circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m
ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)
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